Coll of Exercises II

8/8/2019 Coll of Exercises II

1/52

Collection of Exercises

for the lecture

Electrical Machines II

by

Univ. Prof. Dr.Ing. G. Henneberger

RWTH Aachen

Edited and Processedby Assistants of the Institute for Electrical Machines.

Michael Bork

Andreas Brosse

Babak Fahimi

Thomas Friedrich

Matthias Schmitz

Reprint Prohibited

1. Edition

2003


2/52

Exercise P1

A permenant excited DC machine has the following nominal data:

# % ' 0 2 4 3 0 6 8

The armature inductance amounts to@ A C E

2G F

and the moment of inertia is

A H

P ' S U

. Satu-

ration and all losses except the armature copper losses are to be neglected.

1. Calculate the armature resistanceV W

, the no-load speed0 X

at nominal voltage, the ar-

mature time constant` W

, the nominal starting time` a

and the mechanical time constant

` b of the machine.

2. The machine is coupled with a load machine with a moment of inertia

A

E

' S U

. The

loaded standstill machine is suddenly connected to a 110V power supply. Which speed

appears after a long time? What is the speed of the machine after a time equal to two

times the armature time constants at starting? Guess with the help of damping D, whether

the starting is aperiodic or under oscillations.

3. The DC motor is now supplied from a DC voltage supply of 300 V through a DC current

converter with 1 kHz pulse frequency. Determine the ratio` d

e

` g h p rwhich has to be

chosen, so that at nominal torque, the speed is@ E

R 2 G 3 0

6 8

!

4. The machine is supplied from an ideal current converter, whose permissible armature

current limit is 100A! Further the regulation is limited to a maximum angular ac-

celeration ofH

S

6 t

. Determine the minimum reverse time fromv H

2 3 0

6 8

to

y

H

2 3 0

6 8

and the armature current for the unloaded and loaded case as in b).


3/52

Note:

6 8

H

S

t

X

S

t

v

S

X

v

t

X

H

y " $

&

8 6 ' )

S 3 0 1

X 3

H

y

t

5 v 6 8 9

S 3 0 3

H

y

t A

H

y B

H v

X

5 D G

6 I

" Q

H

y " $

&

' ) 6 8

S

S 3 0 U1

X 3

t

y

H 5

A

v

3

t

y

H

9 W

S U1

X 3

t

y

H 5

A Y

a

W 8

c

H

a

W 8

H

a

W 8

d

H

2


4/52

Solution

1.

V W

y

y

e

y

# % 'e

V W

A

#

# % '

E A H

0

0 X

0

E A H

2G 3 0 6 8

0 X

H

2 3 0 6 8

`W

W

V W

@ A C E

2 F

A

#

`W

2 G S

a

t

t

0 X

0

A H

P ' S U

t

t

H

C

S

# % '

C

S

` a

H E

# 2 S

` b

8 W

` a

V W

W

` a

A

#

H E

# 2 S

` b

H

A C

2 S

2.

Q

d

Q

Wp

B

A E v

A H

P

D

' S

U

A C

P ' S

U

` b

0

G ` b

` b

`b

d

p

Q

d

Q

Wp

d

p

H

A @

2G S

A C

P

A H

P

A

@

S

! ` b

P

` "

!

A

@

A

#

H A C

# d

H

6 $ G 8

3

W &

3

9 9

U

6 8 6 9 5 G 8

3 S

5

3

9

5 ( ) 1

0 0 X

0X

0

E

2G 3 0 6 8

X

! H

` W

` b

!

H

A

A

@

H

S

H

P

A

# P S6 8

0

B

5 D

0

B

5 ( ) D

H

y

v

3

t

y

H

3

t

y

H

G

6 I

" 4

' 6

&

' ) 6 8 6

Q

v

y

3

t

y

H

3

t

y

H

G

6 I

" 4

' 8

&

')

6 8 6

Q

3


5/52

H A C

# P

E

t

y

H

H A

E E C

0

B

5

`" D

0

B

5 ( ) D

H

y

A

P

v H

A

C

H

6

U

A

@ E

0

B

5

` " D

P #

A @ @

2 G 3 0 6 8

3.

0

@ E

2 3 0 6 8

0

0 X

@ E

H

H C E

W

v

V "

H C E

v

A

H

H

#

`

e

` g h p r

W

H

#

A C

4. Without loading:

&

&5

&

&5

# 2

A H

P ' SU

Assumption

H

(

H

2

&

&5

H

' S

A H

P ' S

U

H C

A E

H

S

t

d

H

H

S

t

&

&5

&

&5

b W

H

H

S

t

y

H

H C

A E

H

y

@

#

0

v H

2 3 0 6 8

y B y

H

R 2 G 3 0 6 8

D

R 2 G 3 0 6 8

&

0

&5

H

&

&5

H

C

P

A H

H

S 2 G 3 0

5

0

&

0e

&5

A H E

# S

A C

P ' S U

H

H

2

&

& 5

H

' S

A C

P ' S

U

H

S

t

c

H

H

S

t

y

H

&

0

&5

@ C

P

A

H

S 2 3 0

5

A C H

S

4


6/52

Exercise P2

For an universal motor the following rating plate data is known:

a

E

F 0

C

2 G 3 0

6 8

Apart from that, the following data is also measured:

" v

H

2 G F

V "

V

A E

#

H

6

' SU

Saturation and all losses except the copper losses in armature and excitation winding are to be

neglected.

1. Determine the nominal power, the nominal power factor, the nominal efficiency and the

nominal torque.

2. Which DC current and DC voltage must be adjusted so that the universal motor at a DC

voltage supply can be operated at the nominal speed with the same thermal loading of

the windings? Which rotational torque is given by it? How is the magnetic utilization

campared to a)?

3. The series winding is replaced by a permanent magnet, so that the flux as under b)is

obtained.The total inductivity of the machine is therefore much reduced to 0,3 times

the measured value.How big is the no-load speed at 220V DC voltage? Determine the

armature time constant, the nominal starting time and the mechanical time constant. The

nominal current determined under b) is essential.

4. Determine time characteristic of the speed while the machine after c) is running at

no-load from standstill, when an ideal armature current converter is adjusted to 4 times

the nominal current? After which time it will reach 1000 rotations/minute? How much

kinetic energy is stored at the end of the run-up? Which compensation capacitor is

equivalent to the rotating armature?

5


7/52

Solution

a)

E

H

2 F

H A

P

H

@

H A

P

H

P

A

V

V " v V

B

V " v V

D

E

H E

B

v V

D

t

v

B

D

t

t

t

y B

D

t

y

V

H

# #

H

#

A

#

E E H A

P '

S 3 0

P

A

E A

P

X

9 W

S

A

H

#

A

#

A

A

#

E E H A

P '

C

e

C

S

A

# #

2

b)

Equal thermal loading: I=

=3A (same effective value)

0 0

B

G8 9

3 S

G D 8

G

2

6

35 0 S S

6

2

G

A

# #

2

6


8/52

Magnetic utilization:8

&

t

H

#

A

#

"

E

v H

#

A

#

H

#

A

#

c)

H

2 F

A

2 F

from b):

0

H

#

A

#

C

e

C

S

H A

# # S

9

0

0 X

0

0 X

0 X

C

e

2 3 0

H

#

A

#

@ H

# e

2 3 0

`"

V "

2 F

A E

H

2 S

Nominal starting time:

` a

0 X

#

H

6

' S U

@ H

# e 2 3 0

A

# #

2

C

#

A @

2 S

mechanical time constant:

` b

` a

V "

e

` b

C

#

A @

A E

e

A

2 4 S

7


9/52

d)

linear:

0

B

5 D

0 X 5

` a

5

8

H

@ H

#

` a

P

H

@ H

#

C

#

A @

2 S

H

P

A

# 2 S

'

H

B

0 X

D

t

t 0 t

X

t

B

@ H

#

C

SD

t #

H

6 ' S

U

C A

' S

'

b W

'

H

9

t

9

C

B

D

t

A

E

H

6

U

8


10/52

Exercise P3

An 8-pole salient pole synchronous machine has the following nominal data:

P % #

(line to line)

9 W

S

A @

B

D

a

C

F

In the test field, the three phases of the stator of an unexcited machine is supplied with nominal

frequency and the machine is driven with a slip of 0,1%. Therefore the following current and

voltage (line to line) are measured:

b W

P

b

s

C

b W

P

E

b

s

The effective value of the sudden short circuit current at half the no-load excitation and nominal

frequency amounts to

C

. A damper winding is not existing. The no-load excitation

current amounts to

X

E

. The short-circuit time constant amounts to`

A

S

G9

.

Saturation and all losses are to be neglected.

1. Determine the per unit reactances

,

and

! Which value has the total leakage ratio ?

2. Determine the displacement angle, the field current and the working torque of the turbine

at nominal operation.

3. The machine is loaded suddenly at the nominal voltage and nominal speed, from no-load

with star connected reactances whose phase value are equal to the d-axis reactances of

the salient pole machine. Determine the terminal voltage after switching and after a long

time?

4. The machine is operated at continuous short circuit with no-load excitation.Suddenly the

short-circuit is switched off. Calculate the terminal voltage immediately after switching

and after a long time. Determine the time characteristic of the effective value of the

terminal voltage.

9


11/52

Solution:

1. With

t

B

#

D

t

P %

A C

we get:

b W

e

3

b

P e

3

A @ @

(

H A

#

b

e3

b W

C

e3

P

E

A

P

C

(

A @ @

Sudden short-circuit experiment at half-the no-load excitation (

g

t

&

U

):

e

3

#

3

C

A H

#

(

A

A

H A

#

A

P

2. For the nominal operation it is 3

H :

5 6

0

3

9 W

S

v

3

S 3 0

A @ @

A @

H v

A @ @

A @ H

A

P #

(

H

A

d

With

H

A

d

v

P

E A C

d

C

P

A

#

d

:

g

9 W

S

v

3

S 3 0

H

9 W

S

H

A

d

v H

H A

#

S 3 0

C

P

A

#

d

A H

(

g

X

H

E

Nominal torque:

$

9 W

S

a

P

P %

A @

C

S

6 8

H @

#

2

3. Before Switching:

g

g

H

and

3

After Switching:

,

d

and

.

g

9 W

S

v

3

S 3 0

3

9 W

S

v

3

S 3 0

and therefore:

3

g

9 W

S

v

S 3 0

H

H A

#

H v

A

P

H

A C

3

H A

#

A C

A

# #

After long time:

3

g

9 W

S

v

S 3 0

H

H A

#

v H A

#

A

H (

3

A E

10


12/52


,

,

d

,3

H

e

A @

#

and g

H.

g

9 W

S

v

3 S 3 0

v

A

A @

#

A

P

After Switching:

g

A

P

.

After long time:

g

H .

Exponential transition from 0,234 to 1:

gv

B

g

y

g D G

6

$

"

with`

X

`

e

H A

@

S

.

11


13/52

Exercise P4

For a 16 pole salient pole synchronous machine Ba

r

E

S

6 8

D , the following nominal data is

known:

C A

%

(Line to line)

H

9 W

S

A

#

The displacement angle in nominal operation amounts to

.

When the mechanically unloaded machine is loaded with nominal excitation current, so it sup-

plies an inductive reactive power of

A

# #

into the supply. Further the following

experiments were carried out.

I. The machine runs at no-load and is excited with nominal voltage. Then the excitation

windings are short-circuited and the terminal voltage is oscillographed. The evaluation

of the oscillograms gave the decay time constant of S

.

II. The stator terminals are short-circuited in 3 phases and the rotor is running with synchro-

nous speed. The excitation is so adjusted that nominal current flows. Then the excitation

winding is short-circuited and the delayed stator current is oscillographed. One finds the

decay time constant as 0,544 s.

1. Determine the synchronous reactances of the d and q axis B

A

D !

2. Calculate the no-load excitation current and the transient reactance

.

3. Starting from the nominal operation the machine is suddenly overloaded. Which rotatio-

nal torque is developed, when the displacement angle reachesH

and it is assumed that

the flux linkage of the excitation winding stays constant?

4. Which excitation current is reached under c) when

H

?

12


14/52

Solution:

1.

3

H

3

C A

%

H A C

P

9 W

S

A

#

C A

#

@

v

v

C A

#

@

C C A

#

@

At nominal operation:

3

S 3 0

9 W

S

S 3 0

3

9 W

S

C A

%

S 3 0

3

H A C

P

9 W

S

C C A

#

@

E

A E

The machine operates at no-load with nominal voltage and nominal excitation. i.e.:

3

A

g

g

So

&

U

and therefore:

g

v

3

v

3

e

3

y

e

3

9 W

S

S 3 0

y

e

B

3

D

C A

% e

3

y

C A

% e

3

9 W

S

H A C

P

S 3 0

C C A

#

@

y

A

# #

e

B

3

C A

%

D

C E A H

#

referred to the nominal impedance

)

A C

:

H A

@

H A C

P

2. For the excitation voltage in nominal operation:

g

v

3

9 W

S

v

S 3 0

C A

%

3

9 W

S

v C E A H

#

H A C

P

S 3 0

C C A

#

@

#

C

P

13


15/52

and in no-load:

g

X e3

Also:

X

g

X

g

C

3

#

C

P

H

A E

#

From Experiment I we get:`

X

S

; and from Experiment II:`

A E

P P S

So:

X

`

`

X

A E

P P

A H

#

H

Also:

X

A H

#

H

H A C

P

A

@

3. constant rotor flux linkage

g

9 W

0 S

5

Output state is the nominal operation, Therefore:

g

g

Also it is:

g

g

3

9 W

S

v

S 3 0

C A

%

3

9 W

S

v

A H

#

H C E A H

#

H A C

P

S 3 0

C C A

#

@

P

H

P P

In this case the machine is working in transient state:

Q

W

r

$

g

S 3 0

v

H

y

H

t

S 3 0

#

H

P S

6 8

P

H

P P

C

t

3

A H

#

H

C E A H

#

S 3 0

H

v

H

E

A E

y

H

A H

#

H C E A H

#

C

t

t

S 3 0 P

H H

H

#

2

4. It is essential that:

g

9 W

S

v

with

6

d

r

Therefore we get:

g

9 W

S

v

H

S

g

y

9 W

S

Y

14


16/52

resp. at nominal operation:

g

9 W

S

v

H

S

g

y

9 W

S

Y

Subtracting the last two equations and dividing by

g

we get:

Q

W

r

g

g

H

y

H

y

g

B

9 W

S

y

9 W

S

D

H

y

H

y

A H

#

H

A H

#

H

C

3

#

C

P

B

9 W

S

H

y

9 W

S

D

A C

Also it continues:

Q

W

r

A C

A C

#

C

A

P

Comment: The relation for the excitation current in the transient case can also be found

by equating the transient and stationary torque equations.

15


17/52

Exercise P5

For a salient pole synchronous machine, whose stator windings are star-connected, the follo-

wing data is known:

H E

%

C

B

G8

%

D

E

P a

E

F

$

e

A H C

No-load experiment:0

H

" 2 4 3 0

6 8

;

H E

;

E @ E

(Line to line)

Short-circuit experiment:0

H

2 3 0

6 8

;

H

A E

;

'

E

1. Determine the no-load excitation current X

!

2. Determine

und

!

3. Which breakdown torque is given at

and at which displacement angle it appears?

4. The machine, starting from nominal operation, is disconnected from the supply. Calcu-

late the voltage and the excitation current immediately after the switching!

Solution:

1. From the no-load experiment it follows:

X

0

0

X

H E

C

E @ E

H

H E

H

A

P

2. From the short-circuit experiment:

e

3

'

0

0

E @ E

e3

E

H E

H

H

A E

H E

P

A

@

is found with the help of the vector diagram:

g

X

E

H

A

P

A

P

3

H E

%

3

C

H

P P

A

P

g

y

9 W

S

C

e

3

B

A

P

y

9 W

S P

d

D

P

A

@

H

A

6 8 9

S 3 0

C

P

16


18/52

From the vector diagram one reads:

y

A

P

H H A

A

H

B

H

y

A

HD

A

P

E

3. For

we get:

$

t

H

y

H

S 3 0

From the formula it is given: P

E

and by replacing the value:

C

t

E

H

A

P

E

y

H

P

A

@

2

C

#

A

H

2

4. Immediately after the switching there is only the transient excitation voltage at the ter-

minals. Therefore:

B

5

v

D

g

9 W

S

v

C

e

3

9 W

S P

d

v

A H C

P

A

@

H

A

E E A @ @

17


19/52

Immediately after the switching

g

g

. Therefore it follows for the excitation cur-

rent:

3

B

5

v

D

g

X

E E A @ @

3

C

H

A

P

H

A @

18


20/52

Exercise P6

Given is a salient pole synchronous machine with the following data:

@ E

C A

# %

9 W

S

A

#

$

# a

E

S

6 8

Further it is known that%

A

#

. At no-load at nominal voltage the stability border is found

at the excitation current of 30% of the no-load excitation current. The saturation and all losses

can be neglected.

1. Give

, and the displacement angle at the operation withC

and9 W

S

H

and with the nominal voltage and frequency!

2. Starting from the operating point in a) the machine shall bring a transient torque at

e

B

9 W

0 S

5

D , that has three times the value of the nominal torque. Calculate the

transient reactance!

3. Which overexcited reactive power B 9 WS

D the machine supplies continuously wi-

thout any winding be overloaded?

4. Determine the excitation current in the transient state after b), referred to the nominal

excitation current!

Solution:

1.

H

%

H

A

#

H A

E

Antiexcitation experiment:

2

g

S 3 0

v

H

H

y

H

S 3 0

No-load: g

y

A

;

;

H

&

2

&

g

9 W

S

v

H

y

H

9 W

S

H

y

g

v

H

H

B

H

y

g D

H A

H A

E

H A

P

A

C

19


21/52

$

X

A

#

9 W

S

H

H

3

A

#

56

0

3

9 W

S

H v

3

S 3 0

3

A

C

A

#

A @ C

#

@ A E

2.

Q

W

r

H

9 W

S

g

S 3 0

v

H

H

y

H

S 3 0

H

9 W

S

g

For

g

we get:

g

9 W

S

v

3

S 3 0

9 W

S

@ A E

v

A

# 3

S 3 0

@ A E

9 W

S

A

#

By introducing the determined relations into one another, we get :

9 W

S

@ A E

v

A

#

S 3 0

@ A E

A

#

A @

A

P

y

A

P # #

A

P

H E

3. Condition:3

Hand g

g

g

9 W

S

v

S 3 0

C A

#

@

56

0

9 W

S

H v

S 3 0

A

C

A

#

H v

A

C

A C

A

P #

@

E A

#

g

A

H

overexcited,9 W

S

,

g

9 W

S

v

3

S 3 0

H v

3

g

A

E

d

g

. This case is not permitted, So:

g

g

A

H

3 H A

H

H A

E

A

# #

A

# #

@ E

C

A C

20


22/52

4.

g

9 W

0 S

5

9 W

S

v

3

S 3 0

3

S 3 0

B

v

D

9 W

S

W

6

v

3

W

6

S 3 0

W

6

9 W

S

@ A E

v

A

P

H E

A

#

S 3 0

@ A E

A

E

For

e

also it is needed that:

g

3

S 3 0

g

H A

E

A

P

H E

A

E

A

H

H A

P

21


23/52

Exercise P7

A 30 pole hydraulic driven synchronous generator has the following nominal data:

C

H

A E

%

(Line to line)

9 W

S

A

# a

E

F

d

C

The following measurements are taken from the test field:

0 e 2 3 0

6 8

Line to linee %

I/kA 9 WS e

8 6 &

e

100 0 24,75 600

200 10,5 33 1 45

Except that

A H Eand

`

A

E

S

is known. The effect of the damper winding,

saturation and all losses except the field winding resistance are to be neglected.

1. Calculate the referred reactances

, and

!

2. Determine the displacement angle, the field current and the driving torque of the turbine

in nominal operation.

3. The machine is at no-load,nominal voltage and nominal speed is suddenly loaded with

a star-connection of inductances, whose phase value are just equal to the quadrature

reactances of the salient pole synchronous machine. Determine the terminal voltage and

the current immediately after switching and after a long time.

4. The salient pole synchronous machine is in continuous short-circuit at 1,33 times the no-

load excitation current. Suddenly the short-circuit is switched off. Determine the terminal

voltage directly after the switching and after a long time. Determine the time characteri-

stic of the effective value of the terminal voltage!

22


24/52

Solution:

1. Measurement 1:

d (

Q

B

a

B

0

D D

Q

A

P

E

3

C

3

H

A E

%

%

Q

Q

%

P

A @ E

%

H A

Measurement 2:

9 W

S

H5

6

0

Q

56

0

Q

Q

H

H

H

A H E

H A

A

2.

C A

@

C

H

A

P

9

2

E 9

2

H

E

"

0

9 W

S

0

C

A

#

t

X X

X

r

A

2

H

E 9

2

3.

H

Before switching:

H

3

H

23


25/52

Immediately after switching:

H

v

H

d

9 W

S

v

3

S 3 0

d

9 W

S

v

d

S 3 0

d

9 W

S

v

S 3 0

H

H v

X

t

8

H

A

#

3

d

y

d 9 W

S

S 3 0

H

y

A

#

A

A

#

3

9 W

S

v

S 3 0

H

H

H v H A

H

A

P

3

A

P

4. Continuous short-circuit:

before switching:

3

H A

3

H

3

A

After switching:

d

A

d

A

H

A E

%

3

H A

%

H A

H A

H

A E

%

3

#

A H

%

`

d

`

A

E

S

A H E

H A C C

S

#

A H

%

v

B

H A

y

#

A H D

%

G

6

$

(or sketch with all the known values)

24


26/52

Exercise P8

For a 4-pole induction machine with short-circuited rotor the following rating-plate data is

given:

#

(Line to line)

C

9 W

S

A

# # a

E

F

The breakdown slip amounts toS

g g

A H Eand the ratio

g g

e

A

P

. The nominal

point is not the optimal point. Saturation and all losses except the rotor copper losses are to be

neglected.

1. Determine the nominal torque, the circle radius and the no-load current.

2. Calculate the nominal speed, the efficiency in nominal operation as well as the starting

torque and starting current!

3. The induction motor is supplied by a converter with variable voltage and frequency so

that it results in a constant stator flux linkage which is equal to the no-load flux linkage.

Which stator voltage appears, when the machine is supplied with F

stator frequency.

Calculate the stator current, rotor frequency and speed by loading with 1,5 times nominaltorque.

4. The induction machine is operated by means of a converter through field oriented regu-

lation with constant rotor flux linkage which is equal to the no-load flux linkage. Which

stator frequency has to be adjusted, so that a speed ofC

2 3 0

6 8

appears? Determine

the stator current!

25


27/52

Solution:

1.

0

'

0

d

3

9 W

S

X

r

t

3

#

C

A

# #

X

r

t

#

@

2

t

g g

A

P

9 W

S

A

P

C

A

# #R P #

Drawing a circle diagram with

2

E

"

b

#

A

E

(

d

C E

26


28/52

2.

S

A E

(

C A E

0

H

P P

@ A E

2G 3 06 8

"

H

C

"

#

@

2

9

2

P

A H 9

2

C

@

2

3.

8

8

d

8

d

8

d

d

Q

d

d

d

Q

d

a

const. (2

A

2

remain the same

8

A

P

#

3

H E

3

(Phase value)

a

t

S

a

8

9

2

E A

9

2

A H E

E

F

A H

@

F

A E

P F

0

F

B

H

y

A H

@D

E

A @

2 3 0 6 8

4.

t

t

d

Then we get:

g g

t

g g

H A E

H A E

g g

A

P(From Exercise)

g g

H A E

A

P

g g

t

g g

a

t

H A E

P

A

#

S

g g E

F

A

P F

a

8

C

C

S

v

A

P F

A

P F

d

C E

C E

v

P #

C A H

8

d

H

t

v

a

t

a

t

g g

t

C E

H v

A

P F

A

C H

A H E

E

F

t

#

27


29/52

Exercise P9

A 3 phase salient pole synchronous machine is used as reactive power machine. It has the pole

pair number p = 3, the nominal reactive power

H

B

D , the nominal voltage

H E A @ E kV (Line to line) and the nominal frequencya

E

S

6 8

and the three phase

winding is star connected. It is also known, that

A C E

8

A H E

t

A

E

With (negative) excitation current of 100A the machine is just unstable, it is working practically

exactly at the stability border

X

. It is then taking an inductive reactive power of

90 MVA.

All losses, the saturation as well as the effect of damper winding can be neglected.

1. Calculate g

at the stability border with negative excitation current in terms of

. De-

termine !

2. Determine the excitation current at nominal operating point!

3. Due to a switching operation, subsequently at the nominal operation the voltage has an

angular displacement of

and the amplitude falls to

A

. Determine the rotational

torque in the machine immediately after the switching operation.

4. Which excitation current appears in the case of c) also directly after the switching?

Solution

1.

r

2

g

S 3 0

v

H

t

H

y

H

S 3 0

&

2

&

g

9 W

S

v

t

H

y

H

9 W

S

&

2

&

X

g

v

t

H

y

H

g

H

y

H

3

H

A

g

y

3

H

y

A

H

y

A

H

y

H

A

H A H H

28


30/52

2. Calculation of

:

8

r

A H E

t

A

E

8

v

8

v

8

v

8

v

8

e

v H

8

e

v

e

A H E v H

A H E v

A C E

H A

P

@ E

H A E

C

Reactive power machine:

;

;3 3

;3

.

Nominal operation: g

v

3

H v

3

H v H A E

C

A E

C.

Anti excitation experiment:

g

y

3

H

y

H A E

C

A

y

A

P

C

P

H

g

g

y

H

A E

C

y

A

P

C

P

E

P

A

3.

H

y

H

B

H v

r D

B

H v

D

H

y

H

H A H E

H A

E

A H @

g

v

3

H v

A H @

H A E

C

H

H A

@ E

Q

W

r

r

2

Q

W

r

g

S 3 0

v

H

t

H

y

H

S 3 0

A

H A

@ E

A H @

H A E

C

S 3 0

v

H

B

A

D

t

H

H A H H

y

H

A H @

H A E

C

S 3 0

C

H A H

#

r

$

H

P S

6 8

H

E E

%

2

Q

W

r

2

Q

W

r

r

H A H

#

E E

%

2

Q

W

r

H A

@ @

2

29


31/52

4.

3

g

y

9 W

S

H A

@ E

y

A

9 W

S

A H @

H A E

C

H A

#

E

g

9 W

S

v

3

A

9 W

S

d

v H A E

C

H A

#

E

g

A C C

g

g

E

P

A

A C C

A E

C

# #

A

#

30


32/52

Exercise P10

A salient pole synchronous generator has ata

E

S

6 8

the nominal speedE

2 3 0

6 8

;

with the star connection of the stator winding

H

E

(Line to line);

E

and 9 WS

A

. The excitation current for nominal voltage and frequency at no-load is

110 A. At 90 A and at a speed ofP

E

2 3 0

6 8

the continuous short circuit current is 2025 A.

At the nominal real power and9 W

S

Hand at nominal voltage and frequency the angular

displacement is P

. The no-load time constant of the excitation winding is 3 s, the short-circuit

time constant is 0,85 s.

The saturation, the losses and the effect of damper winding can be neglected.

1. Calculate

, und

!

2. Give the active and reactive power of the machine (under or over excited?) during the

operation with nominal voltage and frequency when the angular displacement isC

and the excitation current amounts to 200 A (This operating point is not continuously

permitted)!

3. Determine the time characteristic of the effective value of the terminal voltage, when the

machine is suddenly disconnected from the supply through the load case given in b)!

4. Give the relative torque developed in the machine (in terms of

) when the machine is

supplied with nominal voltage and frequency at the angular displacement of

if

starting from the operating point after b), the rotor flux remains constant.

Solution:

1.

t

B

H

A E

%

D

t

E

A

E

3

E

3

H

A E

%

@ E

Short-circuit:

H

3

0

X

0

0

0

H

3

X

H

H

A E

%

3

E

H H

H

A

E

H A H H

31


33/52

With`

X

S

und`

A

#

E

S

we get:

`

`

X

A

#

E

A

#

A

#

H A H H

A

H E

From the experiment with9 W

S

Hand pure real power it is needed:

56

0

3

9 W

S

H

56

0 H

A E

%

@ E

A

3

H

A

E

56

0 P

A @ E

2.

g

3

X

H

A E

%

3

H H

H H

%

g

y

8

9 W

S

H H

%

y

H

A E

% e

3

9 W

S

C

H A H H

A

E

E C

8

S 3 0

H

A E

% e3

S 3 0

C

A @ E

A

E

H @ E

t

v

t

E C

t

v

H @ E

t

P

E

P #

6 8 95

6

0

6 8 95

6

0

E C

H @ E

P

E A @

y

P

E A @

y

C

y

H

P

A

3

9 W

S

3

H

A E

%

P

E

P #

9 W

S

B y

H

P

A

D

#

'

3

S 3 0

3

H

A E

%

P

E

P #

S 3 0

B y

H

P

A

D

y

A

P

(underexcited)

32


34/52


g

X

8

9 W

S

v

H

A E

% e

3

9 W

S

C

v

A

H E

A

E

E C

E A

%

After Switching B 5

v

D :

g

X

For 5 ( ) :

g

The no-load time constant is working, also:

g v

B

g

X

y

g D G

6

Q

"

H H

%

v

B

E A

%

y

H H

%

D G

6

Q

U

r

H

A

E

%

y

A

#

@

%

G

6

Q

U

r

(Line to line)

4. The rotor flux linkage is constant.

transient Torque is working!

Q

W

r

e

H

9 W

S

g

S 3 0

v

H

H

y

H

S 3 0

with

und

g

&

U

U

8

X

A

#

@

P

we get:

2

Q

W

r

H

A

A

#

@

P

A

H E

2

Q

W

r

A

#

33


35/52

Exercise P11

A 6-pole induction machine with slip ring rotor and both star connected windings has the

following data:

H

P

E

9 W

S

A

# 0

C

2 G 3 0

6 8

By supplying the stator with nominal frequency and half the nominal voltage one measures,

1) At standstill with open rotor the voltage measured between the slip rings is 250V.

2) At synchronous run a stator current intake of 7,5A.

Saturation and all losses except the rotor copper losses are to be neglected.

a) Determine the nominal slip, nominal power and nominal torque (the nominal point is not

the optimal point).

b) Calculate all the components of the equivalent circuit diagram when the stator stray

reactance vanishes and give the corresponding transformation factor.

c) The machine running at nominal voltage, nominal torque, nominal speed and nominal

frequency is disconnected from the supply. Determine the time charcteristic of the effective

value of the terminal voltage after the switch-off, when the speed remains constant during the

reversal operation and the slip rings are short-circuited.

d) The machine is operated by the short-circuited rotor through field oriented regulation with

constant rotor flux-linkage, which is equal to the no-load flux-linkage. Which frequency

appears by loading with double the breakdown torque? Which stator frequency is adjusted, so

that a speed of 5002 3 0

6 8

is given. Determine the stator current!

34


36/52

Solution

a)

S

H

2 3 0

6 8

y

C

2 3 0

6 8

H

2 G 3 0

6 8

P

3

9 W

S

B

H

y

S

D

3

H

P

E

A

#

A

C

C H A

C

0

C H A

C

C

e

C

S

C H

A

2

b)

35


37/52

8

3 X

H

3

@ A E

#

A E

H A E 9

2

H

P

A

9

2

A H

8

H

y

P

A

#

t

'

y

B

t

H v

8

D

t

V

t

S

B

H v

8

D

t

V

t

B

H v

8

D

t S

3

H

P

E

A

#

y

C H A

C

B

#

D

t

A

P

A E

#

c)

`

t

H

XS

g g

H

A H

X

r

A H

P P

2 S

g

X E A

E A

#

A

X

X

A

0

0X

G

6

$

)

H

3

A

A

C

G

6

$

) )

"

d)

X

g g

t

% 3

$ $

S

g g

aX

a

t

A H

P P

E

F

@ A

F

a

8

E

F

v @ A

F

A

F

8

t

X

v

t

X X

H E

t

v H

P P

t

H

P

E

36


38/52

Exercise P12

For an induction machine with short-circuited rotor all losses except the rotor copper losses

as well as the saturation and current displacement are to be neglected. The machine has 4

poles with star connected stator winding and has a nominal voltage of #

(Line to line).

The machine working at no-load with nominal voltage and nominal frequencya

r

E

S

6 8

has a current of

. Such a no-loaded machine is suddenly disconnected from the supply

through an ideal switch, so that the effective value of the voltage follows this relation with time$

E

G

6

Q

X

r

.

1. Calculate the breakdown torque of the machine at nominal voltage and nominal frequen-

cy!

2. Determine the appropriate break down slip in a)!

3. Give the starting torque of the machine!

4. The no-loaded machine is suddenly short-circuited at the maximum phase voltage. Give

the time characteristic of the current in the relevant phase!

Solution:

1.

B

H

y

D

H

y

H

y

E

#

A

#

From the time characteristics of the voltage it follows:

`

t

X

A E

S

For the real current at the breakdownpoint,

8 6

t

X

. Also:

g g

$

a

8

3

X

H

y

S

E

3

#

H

y

A

#

A

#

@

2

2. For the no-load rotor time constant the following relation is known:

`

t

X H

8

S

g g

Therefore it follows:

S

g g

H

8

`

t

X

H

A

#

E

A E

A

#

37


39/52

3. The kloss formula is needed.Also with s=1:

g g

S

g gv H

eS

g g

@

2

A

#

v H

e

A

#

H H E

2

4. The current can be determined from sudden short-circuit of full pole synchronous machi-

ne. Since the ohmic resistance of the stator winding is neglected, the induction machine

has no continous short-circuit current and the switching instant lies at the voltage maxi-

mum, because of the constant rotor flux-linkage we get :

3

B

5 D

3

e

3

8

8

H

y

G

6

Q

)

"

S 3 0

8

5

3

X

H

y

G

6

Q

)

"

S 3 0

8

5

3

H

y

A

#

A

# G

6

Q

4

X

X

X

r

6

S 3 0

E

S 6 8

5

P # #

G

6

Q

X

X

S 3 0

E

S6 8

5

38


40/52

Exercise P13

For a position drive, an induction motor having the following nominal data is applied:

#

a

E

S 6 8 % ') 0

H

P # 2 3 0 6 8

9 W

S

A

#

The breakdown torque is

A E

. The leakage ratio amounts to 0,088.

1. Determine the breakdown torque at nominal stator flux-linkage as well as the rotor break-

down frequency and the rotor time constant!

With the help of a coordinate transformation, the motor is represented as seperately excited DC

machine. The converter is controlled such that the rotor flux linkage is constant in each opera-

ting state. Therefore the general reference axis A, the transformed winding systems (

,

)

and (

,

) shall coincide with the direction of the transformed rotor flux-linkage.

b) Which transformed DC current

the converter must supply in the stator winding at

no-load, if nominal flux linkage shall exist in the stator?

c) Starting from no-load the machine shall bring up double the nominal torque for a short-

time. Give the currents in the stator winding system

!

d) Give the time characteristics of the current in stator phase u before and after the sudden

load in c) whereas it is assumed that the sudden load occurs in the current maximum of

the phases. The speed is approximately taken as constant ( H 2 3 0

6 8

).

39


41/52

Solution

1.

S

0 X

y

0

0X

H E

y

H

P #

H E

A

H

9 W

S

A

# P

3

9 W

S

B

H

y

S

D

'

3

#

A

#

A

#

@

E A E C

g g

A E

0

A E

% '

H

P # e

C

S

g g

P #

A

P

2

40


42/52

From circle diagram:

S

g g

A

E

S

A

E

A

H

A

C

a

g g

S

g g

a

8

A

C

E

S 6 8

a

g g

A H

S 6 8

`

t

H

a

g g

H

A

# #

A H

S

6 8

`

t

A E

# S

2. magnetising component:

3

"

3

X

8 W

2

1

X

A

3

"

A

#

H

Torque building component:

3

3.

t

`

t

g g(see Exercise FB1)

t

`

t

A E

H

H

H

A E

H

A E

A

# #

P

A E E

3

t

`

t

3

"

P

A E E

A

#

H

H @ A

Rotor flux linkages are kept constant:

3

"

A

#

H

4. Sudden load at 5

.

5

c

:

3

h

3

X

9 W

S

8

5

8

$

0

P S6 8

3

h

B

5 D

3

A

9 W

S

P S 6 8

5

41


43/52

5

:

8

$

0

v

t

P S 6 8

v

P

A E E

A E

# S

P

H A

E

S 6 8

3

!

3

t

v

3

"

t

3

H

A

A phase shift appears which can be calculated directly from the currents in the

two-axis system.

6 8 95

6

0

3

3

"

@ @ A C

A

P

So:3

h

B

5 D

3

H

A

9 W

S

P

H A

E

S6 8

5v

A

P

42


44/52

Exercise P14

An induction motor with the nominal data

#

(line to line)

a

E

S

6 8

9 W

S

A

#

H

%( ' 0

H

P P " 2 G 3 0

6 8

g g

is used as a servomotor and it is supplied such that, the available rotor flux linkage remains

constant at all operating points at no-load with nominal voltage and nominal frequency. The

saturation, the losses and the effect of damper winding can be neglected.

1. Determine the leakage constant and rotor time constant of the motor.

2. For the determination of the supplied current a transformation of the stator winding is

to be carried out, so that the current in rotor magnetising component3

"

and in torque

building component3

can be analysed. Give for the no-load case3

"

3

and determine

the time characteristics of the supplied current in phase u at0

H

2 G 3 0

6 8

(At t = 0

there is maximum flux in phase u).

3. Starting from no-load in b) the motor is brought to the nominal torque. Give in this case

the current in both the transformed winding axes.

4. Give for the loading after c) the time characteristics of the current in the three phases

supplied from the converter! Give a physical explanation, why the effective value of the

current is slightly lower under the operation with constant voltage!

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Solution

1. Solution using circle diagram:

A

S

0 X

y

0

0X

H E

y

H

P P

H E

A

P

B

H

y

S

D

3

9 W

S

H

% '

A

C

3

#

A

#

A

9 W

S

A

A

#

H

g g

#

From circle diagram: X

@ A

#

andS

g g

A H E .

X

X

v

g g

@ A

#

@ A

#

v

#

A

`

t

H

8

S

g g

H

A

E

S

6 8

A H E

`

t

A

S

2. At no load:

3

3

"

3

X

3

@ A

#

3

"

H

A E

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Maximum flux at t = 0:

3

h

3

X

9 W

S

B

$

0

5 D

3

@ A

#

9 W

S

H

C

S6 8

5

3

h

3

@ A

#

9 W

S

A

S6 8

5

3. Magnetisation stays 3

"

3

X

H

A E

g g

t

g g

t

`

t

3

"

3

g g

3

g g `

t

3

"

3

3

"

H

P

A

3

"

A C

H

A E

3

C A

4.

$

0

A

S 6 8

t

3

3

"

H

`

t

A C

H

A

S6 8

H H A C

S6 8

8

v

t

A

S 6 8

v H H A C

S 6 8

E A

S 6 8

3

H

3

3

"

t

v

3

t

3

X

H

t

v

A C

t

3

@ A

#

H

t

v

A C

t

3

A

P

3

h

3

A

P

9 W

S

E A

S 6 8

5v C

A C

3

3

A

P

9 W

S

E A

S 6 8

5

y

E

A

P

3

3

A

P

9 W

S

E A

S 6 8

5

y

H @

A

P

The magnetisation of the machine is succeeded by the field-oriented operation only in

the Aaxis. By the operation with power supply however it will be magnetised in the

direction of A and B axis. Therefore the reactive power demand during the operation is

higher which results in increase in the effective value of current.

45


47/52

Exercise P15

For an induction motor with short-circuited rotor the following nominal data is given:

H

(Line to line)

H

9 W

S

A

#

0

C

R 2 G 3 0

6 8

a

E

S

6 8

At no-load with nominal voltage with a frequency of 60 Hz a no-load current of 33,3 A is

measured.

Saturation, current displacement and all losses except the rotor copper loses are to be neglected.

1. Determine the short-circuit current

8

, the breakdown slipS

g g and the total reactance

8 for the nominal voltage and nominal frequency.

2. Calculate the total leakage ratio , the rotor time constant`

t and the rotor flux linkage

in terms of stator winding number

)

8 8

)

in no-load and in nominal operation!

3. The machine is operated under field-oriented regulation with constant rotor flux linkage

equal to the rotor flux linkage in nominal operation and loaded with 1,5 times the nominal

torque. Which rotor frequency appears? Calculate the stator current.

4. What is the stator voltage and stator frequency, so that by the operation after c) a speed

of 2 3 0

6 8

is attained? Calculate the power factor.

46


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Solution:

1.

X

"

X

"

B

e

a

D

"

B

e

a

D

"

X

"

C

E

A

C

E

P

From the circle diagram:

8

P

A

S

g g

A H

E

8

8

e

3

X

H

e

3

P

H

P

A

P

2.

X

P

P

A H H @

`

t

H

S

g g

8

H

A H H @

A H

E

E

S

6 8

A

S

8

X

8

X

8

8

X H

P

A

P

E

S

6 8

P

H A

# S

t

X

H v

t

B

H

y

D

8

X

B

H

y

D

8

X

B

H

y

A H H @D

H A

# S

H A C

S

From the circle-diagram:

)

8 8

)

)

"

8 8

)

8

t

X

Also:

)

8 8

)

8

t

X

H A C

S

H A

P

C

S

47


49/52

3. From the circle-diagram:

X H

#

X

C

H H

P

A E

Fur

H A E

:

H A E

H @ H A @ E

8

t

v

t

X

H @ H A E

t

v

C

t

a

t

"

a

g g

"

S

g g

a

8

8

U

A H H @

A H

E

E

S

6 8

A @ @

S

6 8

4.a

8

$

0

v

a

t

t

X X

X

r

v

A @ @

S

6 8

H

@ @

S

6 8

Determination of the voltage graphically:

8

8

H H @

H

P

A

P

E

S

6 8

H @ E A E

A

P S

From CD:

8

8

H A

S

8

H A

S

H

A @ @

S 6 8

8

H C

P

A

#

P

H

d

9 W

S

A @ E

48


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Exercise P16

A 4 pole induction machine with slip ring rotor (Y-connection) with the nominal data

#

(Line to line) anda

E

S

6 8

and a no-load current of 75 A is given. The no-loaded

machine is suddenly disconnected from the supply. Immediately after the switching, one mea-

sures an effective voltage of 350 V (Line to line) at the stator terminals, when the rotor current

of

in Phase u of

3

h

H

P

A

C

G

6

Q

X

t

r

in Phase v of3

y

H H

P

A

G

6

Q

X

t

r

and

in Phase w of3

y

C A

C

G

6

Q

X

t

r

flows. The saturation and all losses except the rotor copper losses(exception d)) can be neglec-

ted.

1. Determine the breakdown torque of the machine supplied with nominal voltage!

2. Calculate the breakdown slip corresponding to a).

3. The machine is operated as synchronous machine by supplying DC current of 300 A

between the 2 slip rings.(3rd slip-ring open).Which breakdown torque is developed in

the machine at nominal voltage?

4. During the operation as synchronous machine with supply accross the two slip rings at

no-load,

H

is adjusted. Then the machine is suddenly short-circuited. Give the

time characteristics of the excitation current under the assumptionV

8

A

E

!

49


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Solution:

1.

B

5

v

D

B

H

y

D

H

y

B

5

v

D

H

y

E

#

A

@

X

@ E

A

@

E

g g

y

X

E

y

@ E

P

@ A E

g g

3

g g

0 X

3

#

P

@ A E

H E

S

6 8

e

C

g g

H

# P

2

2.

`

t

H

S

g g

8

S

g g

H

`

t

8

`

t

A

E

S

S

g g

H

A

@

A

E

S

E

S

6 8

S

g g

A H C

3. Considering the flux-linkages (Peak value of the fundamental wave):

t t

$

P

3

provided

g

E

e

3

.

Flux linkage due to DC power supply:

t t

$

P

9 W

S

So:

g

9 W

S

e

3

From

3

h

,

3

,

3

for5

we get:3

H E

50


52/52

so that:

g

E

3

9 W

S

e

H E

P

C C A C @

g g

$

8

g

$

8

X

g

E

S

6 8

@ E

P

C C A C @

g g

C C

#

A E

2

4. With EM II, Gl. 4.93:

3 3 X

H v

H

y

G

6

Q

y

H

y

G

6

Q

9 W

S

5

A

@

`

`

t

A

@

A

E

S

A

S

`

8

8

V

8

8

a

8

V

8

8

a

8

X

V

8

A

@

# e

3

E

S

6 8

@ E

A

E

A

S

3

H

H v

H

y

A

@

A

@

G

6

Q

X

X

t

r

y

H

y

A

@

A

@

G

6

Q

X

X

U

r

9 W

S

B

E

S 6 8

D

3

H

v H

P

G

6

Q

X

X

t

r

y

H

P

G

6

Q

X

X

U

r

9 W

S

B

E

S6 8

D

Coll of Exercises II

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Transcript of Coll of Exercises II